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Show that for $z \in \Bbb{C}\backslash\Bbb{R\_}$ $$ Log(z) = \int_{\langle 1,z \rangle} \frac{dz}{z}, $$ where $Log(z)$ is the principal branch and $\langle 1,z \rangle$ is the line segment from 1 to $z$.

Since $\langle 1,z\rangle$ is a line segment, I thought it was ok to parameterize $\langle 1,z \rangle$ such that $\gamma:[0,1]\to \Bbb{C},\quad \gamma(t) = (1-t) + tz, \quad \gamma'(t) = (z-1), \quad t\in [0,1].$ Then I computed the integral using substitution: $$ \int_{\langle 1,z \rangle} \frac{dz}{z} =(z-1)\int_{0}^1\frac{1}{1-t+tz}dt = \int_{1}^z \frac{1}{u} du$$ But here I got in trouble. Why can't this be done?

Gary
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  • Differentiate both functions. They have equal derivative and coincide at a point, hence they are equal. – Pedro Sep 21 '16 at 21:05
  • So you mean $L(z) = \int_1^z \frac{du}{u}$. It is holomorphic on $Im(z) >0$ and its derivative is $\frac{1}{z}$ so that $L(z) = \log(z) + C$, and since $L(1) = 0$.. Now on $Im(z) < 0$ it is the same. I let you check $Re(z) >0$ and finish – reuns Sep 21 '16 at 21:13

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In THIS ANSWER, I showed that for any rectifiable path $\gamma$ in $\mathbb{C}\setminus\{0\}$ from $1$ to $z=re^{i\theta}$, $0\le \theta<2\pi$, there exist a number $n\in \mathbb{Z}$ such that

$$\bbox[5px,border:2px solid #C0A000]{\int_\gamma \frac{1}{z'}\,dz'=\log(r)+i(\theta+2\pi n)} \tag 1$$

where $n$ is the net number of times $\gamma$ crosses the positive real axis from the fourth quadrant to the first quadrant.

Note that we tacitly assumed that the branch cut was taken along the non-negative real axis, which is not the principal branch.

If we had taken instead the principal branch, then $-\pi < \theta \le \pi$, and the number $n$ would have been equal to the net number of times that $\gamma$ crosses the negative real axis from the second quadrant to the third quadrant.

Therefore, if $\gamma$ is the line segment from $1$ to $z$, $n=0$ and we have

$$\bbox[5px,border:2px solid #C0A000]{\int_\gamma \frac{1}{z'}\,dz'=\log(r)+i\theta}$$

where $-\pi<\theta \le \pi$.

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Mark Viola
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